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I'm trying to calculate $\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{n}k\sin\left(\frac{a}{k}\right)$. Intuitively the answer is $a$, but I can't see any way to show this. Can anyone help? Thanks!

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Is there anything special about $a$? I'm assuming it is an arbitrary real number? – JavaMan Feb 22 '12 at 2:14
Welcome to math.SE! Do you have learned Riemann integration? – leo Feb 22 '12 at 2:14
You may prove a more general statement: If $a_n \to a$, then $\frac{1}{n}\sum_{k=1}^{n}a_k \to a$. This can be proved by $\epsilon-\delta$ argument. – Sangchul Lee Feb 22 '12 at 2:18
a is an arbitrary real number, yeah. And I've learned Riemann integration! – ro44 Feb 22 '12 at 2:27
@sos440: Thanks, I thought I could try something like this. But I'm wondering if there's perhaps a 'special' trick with this particular limit? leo mentions Riemann integration. – ro44 Feb 22 '12 at 2:29
up vote 2 down vote accepted

I’m not going to work out all of the details; rather, I’ll suggest in some detail a way to approach the problem.

First, it suffices to prove the result for $a>0$, since the sine is an odd function. For $a>0$ we have $k\sin\left(\frac{a}k\right)<k\left(\frac{a}k\right)=a$, so $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)<\frac1n\sum_{k=1}^na=a\;;$$ this gives you an upper bound of $a$ on any possible limit.

You know that $\lim\limits_{x\to 0}\frac{\sin x}x=1$, so there is a $c>0$ such that $\sin x>\frac{x}2$ whenever $0<x<c$. This means that $$k\sin\left(\frac{a}k\right)>\frac{a}2$$ whenever $\frac{a}k<c$, i.e., whenever $k>\frac{a}c$. Now suppose that $n$ is very large compared with $\frac{a}c$; then ‘most’ of the terms of $$\frac1n\sum_{k=1}^nk\sin\left(\frac{a}k\right)\tag{1}$$ will be greater than $\frac{a}2$, and hence so will $(1)$ itself. You may have to do a little fiddling to say just how big $n$ should be taken relative to $\frac{a}c$, but it should be clear that this idea works to show that the limit of $(1)$ as $n\to\infty$ must be at least $\frac{a}2$.

But what I did with $\frac12$ can clearly be done with any positive fraction less than $1$: if $0<\epsilon<1$, there is a $c>0$ such that $\sin x>\epsilon x$ whenever $0<x<c$. If you’ve filled in the missing details for the previous paragraph, you shouldn’t have too much trouble generalizing to show that the limit of $(1)$ must be at least $\epsilon a$ for any $\epsilon <1$ and hence must be at least $a$.

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Based on Brian and sos's arguments, I was wondering if I'm correctly generalizing this situation: Say we have $a_n->a$, and let's take some arbitrary $l>0$ (using 'l' instead of epsilon). Then $|(\frac{1}{n}\sum_{k=1}^{n}a_n)-a|=|\frac{1}{n}\sum_{k=1}^{n}(a_n-a)|\leq \frac{1}{n}\sum_{k=1}^{n}|a_n-a|$. For almost every n, we have that $|a_n-a|<l$, yielding that $\frac{1}{n}\sum_{k=1}^{n}|a_n-a|<l$ for almost every n. So $\frac{1}{n}\sum_{k=1}^{n}a_n \rightarrow a$. Does this sound correct? Thanks! – ro44 Feb 22 '12 at 2:49
@ro44: Not quite. You want $|(\frac1n\sum_{k=1}^na_k)-a|=|\frac1n\sum_{k=1}^n(a_k-a)|\le\frac1n\sum_{k=1}^n‌​|a_k-a|$. Then you want to say that for sufficiently large $n$, ‘most’ terms of this sum are less than $l$, so the sum is less than (say) $2l$. There’s still a bit of work to be done show that if $n$ is large enough, the early ‘bad’ terms are a small enough fraction of the total not to mess things up. – Brian M. Scott Feb 22 '12 at 3:37

$\displaystyle \sin x \leq x $ for $x \geq 0.$ Integrating this over $[0,t]$ gives $$ -\cos t +1 \leq \frac{t^2}{2} . $$

Integrating both sides again from $[0,x]$ gives $$ -\sin x + x \leq \frac{x^3}{6} .$$

Thus, $$ x - \frac{x^3}{6} \leq \sin x \leq x.$$

Hence, $$ \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} - \frac{a^3}{6k^3} \right ) \leq \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \leq \frac{1}{n} \sum_{k=1}^{n} k \left( \frac{a}{k} \right). $$

Since $ \displaystyle \sum_{k=1}^n \frac{a^3}{6k^3} $ is convergent, the Squeeze theorem shows that $\displaystyle \frac{1}{n} \sum_{k=1}^{n} k \sin \left( \frac{a}{j} \right) \to a.$

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